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Is symmetry a necessary condition for positive (or negative) definiteness?

If not:

It can be proved that if $\mathbf{A}:(m\times m)$ is a square (non-symmetric) matrix, then $$ \mathbf{z'Az=z'Bz},~~\mathbf{B=B'=(A+A')/2} $$

On the other hand, a positive definite matrix is a symmetric matrix for which:

$$\mathbf{z'Bz}>0,~~ \mathbf{z\ne o}$$

Can we imply that $\mathbf{A}$ which is a non-symmetric matrix, is positive definite?

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up vote 1 down vote accepted

It looks like you are working with real matrices.

Most often the definition os positive definite includes symmetric, but sometimes this is not required.

In any case, if $z'Az\geq0$ then $z'A'z=(z'Az)'=z'Az\geq0$. So $$ z'Az=\frac12(z'Az+z'A'z)=z'\left(\frac{A+A'}2\right)z. $$

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So, This means that if $\mathbf{z'Az}\geq0$, whether $A = A'$ or not, we call $\mathbf{A}$ a Positive definite Matrix? – Ramin Dec 9 '12 at 7:48
I wouldn't do it, but many people does. – Martin Argerami Dec 9 '12 at 10:30
You mean I can find a reference that defines Positive definiteness for non-symmetric matrices? – Ramin Dec 9 '12 at 14:00
The thing is that when you work with a complex vector space, you get Hermitian for free; but not in the real case. This issue with the definition is mentioned in the wikipedia article:… but I'm not familiar with references dealing with positive definite matrices. – Martin Argerami Dec 9 '12 at 14:48

Yes, it is possible. In fact, it follows easily from the properties of taking the transpose:

$$0 < z^{\mathrm{T}}\left(A+A^{\mathrm{T}}\right) z = z^{\mathrm{T}}Az + z^{\mathrm{T}}A^{\mathrm{T}}z = z^{\mathrm{T}}Az + (z^{\mathrm{T}}Az)^{\mathrm{T}} = 2z^{\mathrm{T}}Az$$

Taking the transpose of a real number doesn't change anything. Therefore $z^{\mathrm{T}}Az> 0$ if and only if $z^{\mathrm{T}}Bz>0$.

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Would you please see my comment at Martin Argerami answer. Thanks – Ramin Dec 9 '12 at 7:50

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